Question: You have found the following ages (in years) of all 5 porcupines at your local zoo: $ 21,\enspace 12,\enspace 14,\enspace 25,\enspace 8$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{21 + 12 + 14 + 25 + 8}{{5}} = {16\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $21$ years $5$ years $25$ years $^2$ $12$ years $-4$ years $16$ years $^2$ $14$ years $-2$ years $4$ years $^2$ $25$ years $9$ years $81$ years $^2$ $8$ years $-8$ years $64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{25} + {16} + {4} + {81} + {64}} {{5}} $ $ {\sigma^2} = \dfrac{{190}}{{5}} = {38\text{ years}^2} $ The average porcupine at the zoo is 16 years old. The population variance is 38 years $^2$.